Mathematics of Cryptography Part I

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1 CHAPTER 2 Mathematics of Crptograph Part I (Solution to Practice Set) Review Questions 1. The set of integers is Z. It contains all integral numbers from negative infinit to positive infinit. The set of residues modulo n is Z n. It contains integers from 0 to n 1. The set Z has non-negative (positive and zero) and negative integers; the set Z n has onl non-negative integers. To map a nonnegative integer from Z to Z n, we need to divide the integer b n and use the remainder; to map a negative integer from Z to Z n, we need to repeatedl add n to the integer to move it to the range 0 to n We mentioned four properties: Propert 1: if a 1, then a ±1. Propert 2: if a b and b a, then a ±b. Propert 3: if a b and b c, then a c. Propert 4: if a b and a c, then a (m b n c), where m and n are arbitrar integers. 3. The number 1 is an integer with onl one divisor, itself. A prime has onl two divisors: 1 and itself. For eample, the prime 7 has onl two divisor 7 and 1. A composite has more than two divisors. For eample, the composite 42 has several divisors: 1, 2, 3, 6, 7, 14, 21, and The greatest common divisor of two positive integers, gcd (a, b), is the largest positive integer that divides both a and b. The Euclidean algorithm can find the greatest common divisor of two positive integers. 5. A linear Diophantine equation of two variables is of the form a b c. We need to find integer values for and that satisf the equation. This tpe of equation has either no solution or an infinite number of solutions. Let d gcd (a, b). If d does not divide c then the equation have no solitons. If d divides c, then we have an infinite number of solutions. One /. of them is called the particular solution; the rest, are called the general solutions. 1

2 2 6. The modulo operator takes an integer a from the set Z and a positive modulus n. The operator creates a nonnegative residue, which is the remainder of dividing a b n. We mentioned three properties for the modulo operator: First: (a b) mod n [(a mod n) (b mod n)] mod n Second: (a b) mod n [(a mod n) (b mod n)] mod n Third: (a b) mod n [(a mod n) (b mod n)] mod n 7. A residue class [a] is the set of integers congruent modulo n. It is the set of all integers such that a (mod n). In each set, there is one element called the least (nonnegative) residue. The set of all of these least residues is Z n. 8. The set Z n is the set of all positive integer between 0 and n 1. The set Z n is the set of all integers between 0 and n 1 that are relativel prime to n. Each element in Z n has an additive inverse; each element in Z n has a multiplicative inverse. The etended Euclidean algorithm is used to find the multiplicative inverses in Z n. 9. A matri is a rectangular arra of l m elements, in which l is the number of rows and m is the number of columns. If a matri has onl one row (l 1), it is called a row matri; if it has onl one column (m 1), it is called a column matri. A square matri is a matri with the same number of rows and columns (l m). The determinant of a square matri A is a scalar defined in linear algebra. The multiplicative inverse of a square matri eists onl if its determinant has a multiplicative inverse in the corresponding set. 10. A linear equation is an equation in which the power of each variable is 1. A linear congruence equation is a linear equation in which calculations are done modulo n. An equation of tpe a b (mod n) can be solved b finding the multiplicative inverse of a. A set of linear equations can be solved b finding the multiplicative inverse of a matri. Eercises 11. a. It is false because b. It is true because c. It is true because 127 is a prime. d. It is true because e. It is false because f. It is false because 8 is greater than

3 3 a. gcd (88, 220) 44, as shown in the following table: q r 1 r 2 r b. gcd (300, 42) 6, as shown in the following table: q r 1 r 2 r c. gcd (24, 320) 8, as shown in the following table: q r 1 r 2 r d. gcd (401, 700) 1 (coprime), as shown in the following table: q r 1 r 2 r a. gcd (a, b, 16) gcd (gcd (a, b), 16) gcd (24, 16) 8 b. gcd (a, b, c, 16) gcd (gcd (a, b, c), 16) gcd (12, 16) 4 c. gcd (200, 180, 450) gcd (gcd (200, 180), 450) gcd (20, 450) 10 d. gcd (200, 180, 450, 600) gcd (gcd (200, 180, 450), 600) gcd (10, 600) 10

4 4 14. a. gcd (2n 1, n) gcd (n, 1) 1 b. gcd (201, 100) gcd ( , 100) 1 gcd (81, 40) gcd (2 40 1, 40) 1 gcd (501, 250) gcd ( , 250) a. gcd (3n 1, 2n 1) gcd (2n 1, n) 1 b. gcd (301, 201) gcd ( , ) 1 gcd (121, 81) gcd (3 40 1, ) a. We use the following table: q r 1 r 2 r s 1 s 2 s t 1 t 2 t gcd s t gcd (4, 7) 1 (4)(2) (7)( 1) 1 b. We use the following table: q r 1 r 2 r s 1 s 2 s t 1 t 2 t gcd s t gcd (291, 42) 3 (291)( 1) (42)(7) 3

5 5 c. We use the following table: q r 1 r 2 r s 1 s 2 s t 1 t 2 t gcd s t gcd (84, 320) 4 (84)( 19) (320)(5) 4 d. We use the following table: q r 1 r 2 r s 1 s 2 s t 1 t 2 t gcd s t gcd (400, 60) 20 (400)( 1) (60)(7) a. 22 mod 7 1 b. 291 mod c. 84 mod d. 400 mod a. ( ) mod 10 (273 mod mod 10) mod 10 (3 7) mod 10 0 mod 10 b. ( ) mod 10 (4223 mod mod 10) mod 10 (3 3) mod 10 6 mod 10 c. ( ) mod 12 (148 mod mod 12) mod 12 (4 8) mod 12 0 mod 12 d. ( ) mod 12 (2467 mod mod 12) mod 12 (7 5) mod 12 0 mod 12

6 6 19. a. (125 45) mod 10 (125 mod mod 10) mod 10 (5 5) mod 10 5 mod 10 b. (424 32) mod 10 (424 mod mod 10) mod 10 (4 2) mod 10 8 mod 10 c. (144 34) mod 10 (144 mod mod 10) mod 10 (4 4) mod 10 6 mod 10 d. (221 23) mod 10 (221 mod mod 10) mod 10 (1 3) mod 10 3 mod a. a mod 10 (a n 10 n a a 0 ) mod 10 [(a n 10 n ) mod 10 (a ) mod 10 a 0 mod 10] mod 10 [0 0 a 0 mod 10] a 0 mod 10 b. a mod 100 (a n 10 n a a 0 ) mod 10 [(a n 10 n ) mod 100 (a ) mod 100 a 0 mod 10] mod 10 [0 0 (a ) mod 100 a 0 mod 100] (a ) mod 100 a 0 mod 100 [a a 0 ] mod 100. c. Similarl a mod 1000 [a a a 0 ] mod a mod 5 (a n 10 n a a 0 ) mod 5 [(a n 10 n ) mod 5 (a ) mod 5 a 0 mod 5] mod 5 [0 0 a 0 mod 5] a 0 mod a mod 2 (a n 10 n a a 0 ) mod 2 [(a n 10 n ) mod 2 (a ) mod 2 a 0 mod 2] mod 2 [0 0 a 0 mod 2] a 0 mod a mod 4 (a n 10 n a a 0 ) mod 4 [(a n 10 n ) mod 4 (a ) mod 4 a 0 mod 4] mod 4 [0 0 (a ) mod 4 a 0 mod 4] (a a 0 ) mod a mod 8 (a n 10 n a a a 0 ) mod 8 [(a n 10 n ) mod 8 (a ) mod 8 (a ) mod 8 a 0 mod 8] mod 8 [0 0 (a ) mod 8 (a ) mod 8 a 0 mod 8] (a a a 0 ) mod a mod 9 (a n 10 n a a 0 ) mod 9 [(a n 10 n ) mod 9 (a ) mod 9 a 0 mod 9] mod 9 (a n a 1 a 0 ) mod a mod 7 (a n 10 n a a 0 ) mod 7 [(a n 10 n ) mod 7 (a ) mod 7 a 0 mod 7] mod 7 a 5 ( 2) a 4 ( 3) a 3 ( 1) a 2 (2) a 1 (3) a 0 (1)] mod 7 For eample, mod 13 [( 1)6 (2)3 (1)1 ( 2)4 ( 3)5 ( 1)3 (2)6 (3)7 (1)2] mod 7 3 mod 7

7 7 27. a mod 11 (a n 10 n a a 0 ) mod 11 [(a n 10 n ) mod 11 (a ) mod 11 a 0 mod 11] mod 11 a 3 ( 1) a 2 (1) a 1 ( 1) a 0 (1)] mod 11 For eample, mod 11 [(1)6 ( 1)3 (1)1 ( 1)4 (1)5 ( 1)3 (1)6 ( 1)7 (1)2] mod 11 8 mod 11 5 mod a mod 13 (a n 10 n a a 0 ) mod 13 [(a n 10 n ) mod 13 (a ) mod 13 a 0 mod 13] mod 13 a 5 (4) a 4 (3) a 3 ( 1) a 2 ( 4) a 1 ( 3) a 0 (1)] mod For eample, mod 13 [( 4)6 ( 3)3 (1)1 (4)4 (3)5 ( 1)3 ( 4)6 ( 3)7 (1)2] mod 13 3 mod 13 a. (A N) mod 26 (0 13) mod mod 26 N b. (A 6) mod 26 (0 6) mod 26 6 mod 26 G c. (Y 5) mod 26 (24 5) mod mod 26 T d. (C 10) mod 26 (2 10) mod 26 8 mod mod 26 S 30. (0, 0), (1, 19), (2, 18), (3, 17), (4, 16), (5, 15), (6, 14), (7, 13), (8, 12), (9, 11), (10, 10) 31. (1, 1), (3, 7), (9, 9), (11, 11), (13, 17), (19, 19) 32. a. We use the following table: q r 1 r 2 r t 1 t 2 t gcd t gcd (180, 38) has no inverse in Z 180.

8 8 b. We use the following table: q r 1 r 2 r t 1 t 2 t gcd gcd (180, 7) mod mod mod 180. t c. We use the following table: gcd (180, 132) has no inverse in Z 180. d. We use the following table: q r 1 r 2 r t 1 t 2 t gcd q r 1 r 2 r t 1 t 2 t gcd e. gcd (180, 24) has no inverse in Z 180. t t 33. a. We have a 25, b 10 and c 15. Since d gcd (a, b) 5 divides c, there is an infinite number of solutions. The reduced equation is We solve the equation 5s 2t 1 using the etended Euclidean algorithm to get s 1 and t 2. The particular and general solutions are Particular: General: 0 (c/d) s k 0 (c/d) t k (k is an integer) b. We have a 19, b 13 and c 20. Since d gcd (a, b) 1 and divides c, there is an infinite number of solutions. The reduced equation is We

9 9 solve the equation 19s 13t 1 to get s 2 and t 3. The particular and general solutions are Particular: General: 0 (c/d) s k 0 (c/d) t k (k is an integer) c. We have a 14, b 21 and c 77. Since d gcd (a, b) 7 divides c, there is an infinite number of solutions. The reduced equation is We solve the equation 2s 3t 1 to get s 1 and t 1. The particular and general solutions are Particular: General: 0 (c/d) s k 0 (c/d) t k (k is an integer) d. We have a 40, b 16 and c 88. Since d gcd (a, b) 8 divides c, there is an infinite number of solutions. The reduced equation is We solve the equation 5s 2t 1 to get s 1 and t 2. The particular and general solutions are Particular: General: 0 (c/d) s k 0 (c/d) t k (k is an integer) 34. a. Since gcd (15, 12) 3 and 3 does not divide 13, there is no solution. b. Since gcd (18, 30) 6 and 6 does not divide 20, there is no solution. c. Since gcd (15, 25) 5 and 5 does not divide 69, there is no solution. d. Since gcd (40, 30) 10 and 10 does not divide 98, there is no solution. 35. We have the equation We have a 39, b 15 and c 270. Since d gcd (a, b) 3 divides c, there is an infinite number of solutions. The reduced equation is We solve the equation 13s 5t 1: s 2 and t 5. The particular and general solutions are Particular: General: 0 (c/d) s k 0 (c/d) t k To find an acceptable solution (nonnegative values) for and, we need to start with negative values for k. Two acceptable solutions are k 35 5 and 5 k 36 0 and In each case, we follow three steps discussed in Section 2.4 of the tetbook.

10 10 a. Step 1: a 3, b 4, n 5 d gcd (a, n) 1 Since d divides b, there is onl one solution. Step 2: Reduction: 3 4 (mod 5) Step 3: 0 (3 1 4) (mod 5) 2 b. Step 1: a 4, b 4, n 6 d gcd (a, n) 2 Since d divides b, there are two solutions. Step 2: Reduction: 2 2 (mod 3) Step 3: 0 (2 1 2) (mod 3) / 2 4 c. Step 1: a 9, b 12, n 7 d gcd (a, n) 1 Since d divides b, there is onl one solution. Step 2: Reduction: 9 12 (mod 7) Step 3: 0 (9 1 12) (mod 7) (2 1 5) (mod 7) 4 d. Step 1: a 256, b 442, n 60 d gcd (a, n) 4 Since d does not divide b, there is no solution. 37. a (mod 5) 3 ( 5 4) (mod 5) 3 4 (mod 5) a 3, b 4, n 5 d gcd (a, n) 1 Since d divides b, there is onl one solution. Reduction: 3 4 (mod 5) 0 (3 1 4) (mod 5) 2

11 11 b (mod 6) 4 ( 6 4) (mod 6) 4 4 (mod 6) a 4, b 4, n 6 d gcd (a, n) 2 Since d divides b, there are two solutions. Reduction: 2 2 (mod 3) 0 (2 1 2) (mod 3) / 2 4 c (mod 7) 9 ( 4 12) (mod 7) 9 1 (mod 7) a 9, b 1, n 7 d gcd (a, n) 1 Since d divides b, there is onl one solution. Reduction: 9 1 (mod 7) 0 (9 1 1) (mod 7) 4 d (mod 50) (mod 50) a 232, b 206, n 50 d gcd (a, n) 2 Since d divides b, there are two solutions. Reduction: (mod 25) 16 3 (mod 25) 0 (16 1 3) (mod 25) / a. The result of multipling the first two matrices is a 1 1 matri, as shown below: ( ) mod 16 10

12 12 b. The result of multipling the second two matrices is a 3 3 matri, as shown below: a. The determinant and the inverse of matri A are shown below: 3 0 A det(a) 3 mod (det(a)) 1 7 mod A 1 7 A adj(a) b. Matri B has no inverse because det(b) ( ) mod 2 mod 10, which has no inverse in Z 10. c. The determinant and the inverse of matri C are shown below: C det(c) 3 mod C (det(c)) 1 7 mod 10 In this case, det(c) 3 mod 10; its inverse in Z 10 is 7 mod 10. It can proved that C C 1 I (identit matri). 40. Although we give the general method for ever case using matri multiplication, in cases a and c, there is no need for matri multiplication because the coefficient of (in a) or (in c) is actuall 0 in these two cases. These cases can be solved much easier.

13 13 a. In this particular case, the answer can be found easier because the coefficient of is 0 in the first equation.the solution is shown below: mod 5 2 mod 5 b. The solution is shown below: mod 7 2 mod 7 c. The solution is shown below: mod 7 1 mod 7 d. The solution is shown below: mod 8 1 mod 8

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